1. Inplicit Differentiation Problem

The problem is to prove that $\frac{dy}{dx}=\frac{-1}{(x+1)^2}$when $x\sqrt{y+1}+y\sqrt{x+1}=0$. I tried substituting $x=\tan^2\theta$ and $y=\tan^2\phi$ but then I got stuck. I tried many other things but this was the only notable one.

thanks.

2. Re: Inplicit Differentiation Problem

Do you know about the chain rule?
For example: Solve $\frac{d}{dx}(\sqrt{x+1})$
Let $y=\sqrt{x+1}$ and $u=x+1$

Then: $\frac{dy}{dx}=\frac{du}{dx}\frac{dy}{du}=(1)(\frac {1}{2\sqrt{u}})=\frac{1}{2\sqrt{x+1}}$

3. Re: Inplicit Differentiation Problem

Yes, I know the basic formulae and algebra of derivatives.

4. Re: Inplicit Differentiation Problem

Just in case a picture helps...

... where (key in spoiler) ...

Spoiler:

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

But this is wrapped inside the legs-uncrossed version of...

... the product rule, where, again, straight continuous lines are differentiating downwards with respect to x.

Larger

Solve the bottom row (for dy/dx).

Edit: ... which is easier said than done. Wolfram doesn't have any better ideas: differentiate x sqrt&#40;y&#43;1&#41; &#43; y sqrt&#40;x&#43;1&#41; &#61; 0 with respect to x - Wolfram|Alpha.

Ok, I'm beginning to see the tan^2 idea...

... nah! Was it suggested? What is the context?

5. Re: Inplicit Differentiation Problem

I took the original equation, moved one term to the RHS, squared both sides of the resulting equation and arrived at this quadratic equation ...

$(x+1)y^2 - x^2y - x^2 = 0$

$y = \frac{x^2 \pm \sqrt{x^4 - 4(x+1)(-x^2)}}{2(x+1)}$

which simplifies to either $y = x$ (which, by inspection, does not work in the original equation) or $y = \frac{-x}{x+1}$

the derivative of the second equation is $y' = \frac{-1}{(x+1)^2}$

might be some other way, but I don't see it ...

6. Re: Inplicit Differentiation Problem

Originally Posted by skeeter
might be some other way, but I don't see it ...
You have chosen the best way in this case.

7. Re: Inplicit Differentiation Problem

I agree with FernandoRevilla. this is the best way. thanks again.